If x and y are integers greater than 1, is x a multiple of y ? (1) \small 3y^{2}+7y=x (2) \small x^{2}-x is a multiple of y. statement (1) alone is sufficient, but statement (2) alone is not sufficient. statement (2) alone is sufficient, but statement (1) alone is not sufficient. both statements together are sufficient, but neither statement alone is sufficient. each statement alone is sufficient. statements (1) and (2) together are not sufficient.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Step-by-step explanation:

A multiple of a number is obtained after multiplying the number by an integer.

Here,

x, y are any two integers greater than 1,

(1) We have,

∵ y is an integer ⇒ 3y + 7 is also an integer,

⇒ y × an integer = x

That is, when we multiply y by a number we obtain x,

∴ x is a multiple of y.

Thus, statement (1) ALONE is sufficient.

(2),

I.e.

, where a is an integer,

∵ x and x - 1 are disjoint numbers,

There are three possible cases,

Case 1 : x is multiple of y

Case 2 : (x-1) is a multiple of y,

Case 3 : neither x nor x - 1 are multiple of y but their product is multiple of y,

Thus, statement (2) is not sufficient.

unless you made a typo, the answers are both a and c.

step-by-step explanation:

1.5/3 is .5 whereas 3/9 is .3333333 not proportional

22/33 is .66666 and 4/6 is also .66666 proportional

2/4 is .5 whereas 142/213 is .66666 not proportional

14/7 is 2 and 7/3.5 is also 2 proportional

you could've also checked it by doing this simple division:

if the larger fraction divides with the smaller fraction to equal 1, it is proportional. so let me show you what i mean:

when i divide 22/33 by 4/6 i get 1, so we know it is proportional, and it is the same with 14/7 and 7/3.5. now if i divide 3/9 by 1.5/3 i get .66666 so therefore we know it is not proportional.

there are many ways to do this so i wanted to show you some possible ways.

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step-by-step explanation:

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